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Author Topic: Moles Again...  (Read 4076 times)

Offline ~~~~shreyapril~~~~

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Moles Again...
« on: September 10, 2009, 02:16:59 pm »
I am sure abt his question as my teacher had given me this one it is not like the previous one..
Here it starts
>>>>>>>>>>>>>>>>
9.3g of impure Sodium bicarbonate[NaHCO3] was heated and 2.2g of CO2 was formed.
Calculate % purity of NAHCO3
pls hepl!!!
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Offline astarmathsandphysics

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Re: Moles Again...
« Reply #1 on: September 10, 2009, 02:24:47 pm »
Quote from: shreyapril on September 10, 2009, 02:16:59 pm
I am sure abt his question as my teacher had given me this one it is not like the previous one..
Here it starts
>>>>>>>>>>>>>>>>
9.3g of impure Sodium bicarbonate[NaHCO3] was heated and 2.2g of CO2 was formed.
Calculate % purity of NAHCO3
pls hepl!!!

molar mass or CO2 is 12+2*16=44 so 2.2/44=0.05 moles of CO2 so .05 moles of NAHCO3. 1 mol of NaHCO3 weighs 23+1+60=84 and 0.5 moles weighs 0.05*84=4.2g

Purity =4.2/9.3=4.5=45%
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Offline ~~~~shreyapril~~~~

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Re: Moles Again...
« Reply #2 on: September 10, 2009, 02:32:01 pm »
oops sorry
i typed the question wrong again..
it is 2.24g not 2.2
sry a silly mistake.
cant understand my own handwritting..
but thanx for the steps astar.
they will help me find the ans
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Offline Ghost Of Highbury

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Re: Moles Again...
« Reply #3 on: September 10, 2009, 03:16:18 pm »
Quote from: shreyapril on September 10, 2009, 02:32:01 pm
oops sorry
i typed the question wrong again..
it is 2.24g not 2.2
sry a silly mistake.
cant understand my own handwritting..
but thanx for the steps astar.
they will help me find the ans
doesnt change the answer much

46%
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Offline ~~~~shreyapril~~~~

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Re: Moles Again...
« Reply #4 on: September 10, 2009, 03:17:54 pm »
i got some wierd ans wen i tried it the way a star told me
i got something near to 98.28% purity
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Offline Ghost Of Highbury

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Re: Moles Again...
« Reply #5 on: September 10, 2009, 03:18:55 pm »
Quote from: shreyapril on September 10, 2009, 02:16:59 pm
I am sure abt his question as my teacher had given me this one it is not like the previous one..
Here it starts
>>>>>>>>>>>>>>>>
9.3g of impure Sodium bicarbonate[NaHCO3] was heated and 2.2g of CO2 was formed.
Calculate % purity of NAHCO3
pls hepl!!!
a simpler way of diong this is use ratio proportion

44g of CO2 is produced with - 84g of pure NAHCO3

therefore 2.24g of CO2 should be produced with - 84*2/44 * 2.24 = 8.55

8.55 / 9.2 * 100 = 93%
« Last Edit: September 10, 2009, 03:34:35 pm by eddie_adi619 »
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Offline ~~~~shreyapril~~~~

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Re: Moles Again...
« Reply #6 on: September 10, 2009, 03:27:09 pm »
oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O

k..
so acc to u it will be
168/44*2.24=8.55
so purity wud be 8.55/9.3*100=91.96 [app]
« Last Edit: September 10, 2009, 03:29:41 pm by shreyapril »
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Offline Ghost Of Highbury

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Re: Moles Again...
« Reply #7 on: September 10, 2009, 03:29:28 pm »
Quote from: shreyapril on September 10, 2009, 03:27:09 pm
oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O
oh..missed that..
u shud have ritten the whole equation

astar missed it too..

u just rote CO2

yes its 84*2
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Offline Ghost Of Highbury

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Re: Moles Again...
« Reply #8 on: September 10, 2009, 03:31:44 pm »
Quote from: eddie_adi619 on September 10, 2009, 03:29:28 pm
Quote from: shreyapril on September 10, 2009, 03:27:09 pm
oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O
oh..missed that..
u shud have ritten the whole equation

astar missed it too..

u just rote CO2

yes its 84*2
i'm getting 92%
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Offline Ghost Of Highbury

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Re: Moles Again...
« Reply #9 on: September 10, 2009, 03:32:42 pm »
Quote from: shreyapril on September 10, 2009, 03:27:09 pm
oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O

k..
so acc to u it will be
168/44*2.24=8.55
so purity wud be 8.55/9.3*100=91.96 [app]
yes
thats wat i rote
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Offline Ghost Of Highbury

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Re: Moles Again...
« Reply #10 on: September 10, 2009, 03:33:26 pm »
Quote from: shreyapril on September 10, 2009, 03:17:54 pm
i got some wierd ans wen i tried it the way a star told me
i got something near to 98.28% purity
how ddid u get

98%
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Offline ~~~~shreyapril~~~~

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Re: Moles Again...
« Reply #11 on: September 10, 2009, 03:33:53 pm »
ohh
but i think u had written 46 not 92 if i am not wrong..
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Offline Ghost Of Highbury

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Re: Moles Again...
« Reply #12 on: September 10, 2009, 03:35:03 pm »
Quote from: eddie_adi619 on September 10, 2009, 03:31:44 pm
Quote from: eddie_adi619 on September 10, 2009, 03:29:28 pm
Quote from: shreyapril on September 10, 2009, 03:27:09 pm
oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O
oh..missed that..
u shud have ritten the whole equation

astar missed it too..

u just rote CO2

yes its 84*2
i'm getting 92%
chk this
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Offline ~~~~shreyapril~~~~

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Re: Moles Again...
« Reply #13 on: September 10, 2009, 03:36:23 pm »
Mr of NAHCO3 is 168         Mr of CO2 is 44
Mass of NAHCO3 is 9.3        Mass of Na2CO3 is 5.9    mass of CO2 is 2.24        mass of H2O is 1
                5.9+1+2.24=9.14
        9.14/9.3*100=98.28[app]
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Offline astarmathsandphysics

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Re: Moles Again...
« Reply #14 on: September 10, 2009, 03:38:49 pm »
i THOUGHT THE EQUATION WOULD BE NaOH3=NaOH+CO2

Which make the ratio 1:1
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